POJ-2155-Matrix(2維樹狀數組)

Description

Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1 <= i, j <= N).

We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a '0' then change it into '1' otherwise change it into '0'). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions.

1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2).
2. Q x y (1 <= x, y <= n) querys A[x, y].

Input

The first line of the input is an integer X (X <= 10) representing the number of test cases. The following X blocks each represents a test case.

The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2 y2", which has been described above.

Output

For each querying output one line, which has an integer representing A[x, y].

There is a blank line between every two continuous test cases.

Sample Input

1
2 10
C 2 1 2 2
Q 2 2
C 2 1 2 1
Q 1 1
C 1 1 2 1
C 1 2 1 2
C 1 1 2 2
Q 1 1
C 1 1 2 1
Q 2 1

Sample Output

1
0
0
1

思路：看到別人的代碼才做出來的，感覺這種思想太厲害了，用2維樹狀數組來維護頂點出現的次數，每加入一個矩形，就把四個頂點加進去（注意邊界問題，x2，y2要加一）。查詢的時候就算出（1，1）到（x，y）的頂點的總數，奇數就為1，偶數就是0。

#include 

int sum[1005][1005],n,m;

int lowbit(int x)
{
    return x&-x;
}

void update(int x,int y)
{
    for(int i=x;i<=n;i+=lowbit(i))
    {
        for(int j=y;j<=n;j+=lowbit(j))
        {
            sum[i][j]++;
        }
    }
}


int query(int x,int y)
{
    int t=0;

    for(int i=x;i>0;i-=lowbit(i))
    {
        for(int j=y;j>0;j-=lowbit(j))
        {
            t+=sum[i][j];
        }
    }

    return t&1;
}

int main()
{
    int T,i,j,x1,y1,x2,y2;
    char s[5];

    scanf("%d",&T);

    while(T--)
    {
        scanf("%d%d",&n,&m);

        for(i=1;i<=n;i++) for(j=1;j<=n;j++) sum[i][j]=0;

        while(m--)
        {
            scanf("%s",s);

            if(s[0]=='C')
            {
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);

                update(x1,y1);
                update(x1,y2+1);
                update(x2+1,y2+1);
                update(x2+1,y1);
            }
            else
            {
                scanf("%d%d",&x1,&y1);
                printf("%d\n",query(x1,y1));
            }
        }

        printf("\n");
    }
}